问题描述
public String generateURLSafeToken(String username, char[] password) throws CredentialTokenException {this.tokenValid = false;字符串令牌 = null;if ((username.length() < 1) || (username == null)) {throw new CredentialTokenException("用户名不能为空字符串或 null.");}if ((password.length < 1) || (password == null)) {throw new CredentialTokenException("密码不能为空或 null.");}
<块引用>
我在第 4 行和第 7 行遇到此错误(用户名 == null 和密码 == null)
我的代码中需要这部分.我正在尝试 isEmpty() 而不是 null 但也面临着问题.解决此 SONAR 错误的替代方法或解决方案是什么
总是计算结果为 false 的条件是 username == null 和 password ==空.
我们以username为例.运算符 || 是 短路 意味着它赢了'如果左侧为 true,则不计算右侧.基本上有两种情况:
- 给定的 username 不是 null.条件 username.length() <1 被评估
- 如果结果为true,我们直接返回,进入if分支
- 如果结果是 false,我们会尝试评估 username == null.但是由于给出的 username 不是 null,因此 always 的计算结果为 false.
- 给定的 username 是 null.条件 username.length() <1 被评估.这实际上停在那里:它会抛出一个 NullPointerException 并且不会评估右侧.
因此,您可以看到,无论何时实际评估 username == null 条件,结果始终为 false.这就是 SonarQube 警告告诉您的内容.
这里的解决方案是颠倒您的 2 个条件.考虑拥有
if (username == null || username.length() < 1)
相反.如果您重新开始并检查每个案例,您会注意到没有一个表达式将始终具有相同的结果:
- 给定的 username 不是 null.第一个条件明确评估为 false,第二个条件被评估,可能返回 true 或 false.
- 给定的 username 是 null.第一个条件明确评估为 true 和短路.
public String generateURLSafeToken(String username, char[] password) throws CredentialTokenException { this.tokenValid = false; String token = null; if ((username.length() < 1) || (username == null)) { throw new CredentialTokenException("Username cannot be an empty string or null."); } if ((password.length < 1) || (password == null)) { throw new CredentialTokenException("Password cannot be an empty or null."); }
I am facing this error in line 4 and line 7 (username == null and password == null)
And I need this part in my code. I am trying isEmpty() instead of null but facing problems in that also . What is an alternate way or the solution to fix this SONAR error
The conditions which always evaluates to false are username == null and password == null.
Let's take the example of username. The operator || is short-circuiting meaning it won't evaluate the right hand side if the left hand side is true. Basically, there are 2 cases:
- The username given is not null. The condition username.length() < 1 is evaluated
- If the result is true, we return directly and enter the if branch
- If the result is false, we try to evaluate username == null. But since the username given is not null, this always evaluate to false.
- The username given is null. The condition username.length() < 1 is evaluated. This actually stops right there: it will throw a NullPointerException and will not evaluate the right hand side.
Therefore, you can see that whenever the username == null condition was actually evaluated, the result was always false. This is what the SonarQube warning is telling you.
The solution here is to reverse your 2 conditions. Consider having
if (username == null || username.length() < 1)
instead. If you start over and go through each case, you'll notice that none of the expressions will always have the same result:
- The username given is not null. First condition clearly evaluates to false and the second is evaluated, which may return true or false.
- The username given is null. The first condition clearly evaluated to true and short-circuits.